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\begin{document}

\title{Complex Analysis}
\subtitle{Chapter 2. Complex Functions \\ 
Section 2. Elementary Theory of Power Series}
%\institute{SLUC}
\author{LVA}
%\date
%\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
%\date{ {2023年9月21日} }

\maketitle

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\begin{frame}{Contents 1-2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item  Introduction to the Concept of Analytic Function
\begin{enumerate}
\item[1.1.] Limits and Continuity
\item[1.2.] Analytic Functions
\item[1.3.] Polynomials
\item[1.4.] Rational Functions
\end{enumerate}

\item {\color{red}Elementary Theory of Power Series}
\begin{enumerate}
\item[2.1.] {\color{red}Sequences}
\item[2.2.] {\color{red}Series}
\item[2.3.] {\color{red}Uniform Convergence}
\item[2.4.] {\color{red}Power Series}
\item[2.5.] {\color{red}Abel's Limit Theorem}
\end{enumerate}

\end{enumerate}

\end{frame}

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\begin{frame}{Contents 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}

\item[3.] The Exponential and Trigonometric Functions
\begin{enumerate}
\item[3.1.] The Exponential
\item[3.2.] The Trigonometric Functions
\item[3.3.] The Periodicity
\item[3.4.] The Logarithm
\end{enumerate}

\end{itemize}

\end{frame}

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\setcounter{equation}{14} % 因为设置为14，下一个使用的标签就会从15开始


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\begin{frame}{2. Elementary Theory of Power Series }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
Polynomials and rational functions are very special analytic functions. 

\item[2.] 
{\color{red}The easiest way to achieve greater variety is to form limits. } 

\item[3.] 
For instance, the sum of a convergent series is such a limit. 

\item[4.] 
If the terms are functions of a variable, so is the sum, and if the terms are analytic functions, chances are good that the sum will also be analytic. 

\end{enumerate}

\end{frame}


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\begin{frame}{2. Elementary Theory of Power Series }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
{\color{red}Of all series with analytic terms the power series with complex coefficients are the simplest. }

\item[6.] 
In this section we study only the most elementary properties of power series. 

\item[7.] 
A strong motivation for taking up this study when we are not yet equipped to prove the most general properties (those that depend on integration) is that we need power series to construct the exponential function (Sec. 3).

\end{enumerate}

\end{frame}

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\begin{frame}{2.1. Sequences. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}\itemsep1em 

\item[1.] 
{\color{red}The sequence $\{a_n\}_1^\infty$ has the limit $A$ if to every $\varepsilon > 0$ there exists an $n_0$ such that $|a_n - A| < \varepsilon$ for $n \ge n_0$. }

\item[2.] 
A sequence with a finite limit is said to be {\color{blue}convergent}, and any sequence which does not converge is {\color{blue}divergent}. 

\item[3.] 
If $\lim\limits_{n\to \infty} a_n = \infty$, the sequence may be said to {\color{blue}diverge to infinity}.


\end{enumerate}

\end{frame}

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\begin{frame}{2.1. Sequences. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[4.] 
Only in rare cases can the convergence be proved by exhibiting the limit, so it is extremely important to make use of a method that permits proof of the existence of a limit even when it cannot be determined explicitly. 

\item[5.] 
The test that serves this purpose bears the name of Cauchy. 

\item[6.] 
{\color{red}
A sequence will be called fundamental, or a Cauchy sequence, if it satisfies the following condition: 

\vspace{0.2cm}

\begin{center}
\fbox{\parbox{10cm}{
given any $\varepsilon > 0$ there exists an $n_0$ such that $|a_n - a_m| < \varepsilon$ whenever $n\ge n_0$ and $m \ge n_0$. 
}}
\end{center}
}

\vspace{0.2cm}

\item[7.] 
The test reads:
A sequence is convergent if and only if it is a Cauchy sequence. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.1. Sequences. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[8.] 
The necessity is immediate. 

\item[9.] 
If $a_n\to A$ we can find $n_0$ such that $|a_n -A| < \varepsilon/2$ for $n \ge n_0$. 

\item[10.] 
For $m,n \ge n_0$ it follows by the triangle inequality that 
\begin{equation*}
|a_n - a_m| \le |a_n - A| + |a_m - A| < \varepsilon.
%\label{eq-}
\end{equation*}

\item[11.] 
The sufficiency is closely connected with the definition of real numbers, and one way in which real numbers can be introduced is indeed to {\color{blue}postulate} the sufficiency of Cauchy's condition. 

\end{enumerate}

\vfill 

\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

{\footnotesize  
postulate: to suggest a theory, idea, etc. as a basic principle from which a further idea is formed or developed
}

\end{frame}

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\begin{frame}{2.1. Sequences. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[12.] 
However, we wish to use only the property that every bounded monotone sequence of real numbers has a limit.

\item[13.] 
The real and imaginary parts of a Cauchy sequence are again Cauchy sequences, and if they converge, so does the original sequence. 

\item[14.] 
For this reason we need to prove the sufficiency only for real sequences.

\item[15.] 
We use the opportunity to recall the notions of {\color{blue}limes superior} and {\color{blue}limes inferior}. 

\end{enumerate}

\vfill 

\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

{\footnotesize  
limes superior: the upper limit \\
limes inferior: the lower limit 

}

\end{frame}

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\begin{frame}{2.1. Sequences. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[16.] 
Given a real sequence $\{\alpha_n\}_1^\infty$ we shall set $a_n = \max \{\alpha_1,\cdots, \alpha_n\}$, that is, $a_n$ is the greatest of the numbers $\alpha_1,\cdots, \alpha_n$. 

\item[17.] 
The sequence $\{a_n\}_1^\infty$ is nondecreasing; hence it has a limit $A_1$ which is finite or equal to $+\infty$. 

\item[18.] 
The number $A_1$ is known as the {\color{blue}least upper bound} or {\color{blue}supremum} (l.u.b. or sup) of the numbers $\alpha_n$; indeed, it is the least number which is $\ge$ all $\alpha_n$. 

\item[19.] 
Construct in the same way the {\color{blue}least upper bound} $A_k$ of the sequence $\{\alpha_n\}_k^\infty$ obtained from the original sequence by deleting $\alpha_1,\cdots,\alpha_{k-1}$. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.1. Sequences. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[20.] 
It is clear that $\{A_k\}$ is a nonincreasing sequence, and we denote its limit by $A$. 

\item[21.] 
It may be finite, $+\infty$, or $-\infty$. 

\item[22.] 
In any case we write 
\begin{equation*}
A = \underset{n\to\infty}{\mathrm{limsup}}\, \alpha_n. 
%\label{eq-}
\end{equation*}

\end{enumerate}

\vfill 

\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

{\footnotesize  

\begin{equation*}
\begin{aligned}
\underset{n\to\infty}{\mathrm{limsup}}\, \alpha_n
& = \inf\limits_{n\in\mathbb{N}} \sup\limits_{k\ge n} \alpha_k \\ 
& = \inf\limits_{n\in\mathbb{N}} \sup \{\alpha_n, \alpha_{n+1}, \cdots, \alpha_{n+p}, \cdots \}.
\end{aligned}
%\label{eq-}
\end{equation*}

}

\end{frame}

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\begin{frame}{2.1. Sequences. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[23.] 
It is easy to characterize the {\color{blue}limes superior} by its properties. 

\item[24.] 
If $A$ is finite and $\varepsilon > 0$ there exists an $n_0$ such that $A_{n_0} < A + \varepsilon$, and it follows that $\alpha_n \le A_{n_0} < A + \varepsilon$ for $n \ge n_0$. 

\item[25.] 
In the opposite direction, if $\alpha_n \le A - \varepsilon$ for $n \ge n_0$, then $A_{n_0}\le A - \varepsilon$, which is impossible. 

\item[26.] 
In other words, there are arbitrarily large $n$ for which $\alpha_n>A-\varepsilon$. 

\item[27.] 
If $A = +\infty$ there are arbitrarily large $\alpha_n$, and $A = -\infty$ if and only if $\alpha_n$ tends to $-\infty$.

\item[28.] 
In all cases there cannot be more than one number A with these properties. 

\item[29.] 
The {\color{blue}limes inferior} can be defined in the same manner with inequalities reversed. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.1. Sequences. \hfill 作业2B-1 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[30.] 
{\color{red}It is quite clear that the limes inferior and limes superior will be equal if and only if the sequence converges to a finite limit or diverges to $+\infty$ or to $-\infty$. 
}

\item[31.] 
The notations are frequently simplified to $\overline{\lim}$ and $\underline{\lim}$. 

\item[32.] 
The reader should prove the following relations:
\begin{equation*}
\begin{aligned}
& \underline{\lim}\, \alpha_n + \underline{\lim}\, \beta_n 
\le \underline{\lim}\, (\alpha_n + \beta_n)
\le \underline{\lim}\, \alpha_n + \overline{\lim}\, \beta_n \\ 
& \underline{\lim}\, \alpha_n + \overline{\lim}\, \beta_n 
\le \overline{\lim}\, (\alpha_n + \beta_n)
\le \overline{\lim}\, \alpha_n + \overline{\lim}\, \beta_n. 
\end{aligned}
%\label{eq-}
\end{equation*}


\end{enumerate}

\end{frame}

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\begin{frame}{2.1. Sequences. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[33.] 
{\color{red}Now we return to the sufficiency of Cauchy's condition.} 

\item[34.] 
From $|\alpha_n - \alpha_{n_0}| < \varepsilon$ we obtain $|\alpha_n| < |\alpha_{n_0}| + \varepsilon$ for $n \ge n_0$, and it follows that $A = \overline{\lim}\, \alpha_n$ and $a = \underline{\lim}\, \alpha_n$ are both finite. 

\item[35.] 
If $a\neq A$ choose
\begin{equation*}
\varepsilon = \frac{A-a}{3}
%\label{eq-}
\end{equation*}
and determine a corresponding $n_0$.


\item[36.] 
By definition of $a$ and $A$ there exists an $\alpha_n < a + \varepsilon$ and an $\alpha_m > A - \varepsilon$ with $m,n \ge n_0$. 

\item[37.] 
It follows that $A-a=(A-\alpha_m)+(\alpha_m-\alpha_n)+(\alpha_n-a) < 3\varepsilon$,  contrary to the choice of $\varepsilon$. 

\item[38.] 
Hence $a = A$, and the sequence converges. 


\end{enumerate}

\end{frame}

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\begin{frame}{2.2. Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
{\color{red}A very simple application of Cauchy's condition permits us to deduce the convergence of one sequence from that of another. }

\item[2.] 
If it is true that $|b_m - b_n|\le |a_m - a_n|$ for all pairs of subscripts, the sequence $\{ b_n\}$ may be termed a {\color{blue}contraction} of the sequence $\{a_n\}$ (this is not a standard term). 

\item[3.] 
Under this condition, if $\{a_n\}$ is a Cauchy sequence, so is $\{b_n\}$.

\item[4.] 
Hence convergence of $\{a_n\}$ implies convergence of $\{ b_n\}$.

\end{enumerate}

\vfill 

\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

{\footnotesize  
contraction: the fact of something becoming smaller or shorter
}

\end{frame}

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\begin{frame}{2.2. Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
An {\color{blue}infinite series} is a formal infinite sum
\begin{equation}
a_1+a_2+\cdots+a_n+\cdots.
\label{eq-15}
\end{equation}

\item[6.] 
Associated with this series is the sequence of its {\color{blue}partial sums} 
$$
s_n = a_1+a_2+\cdots+a_n.
$$

\item[7.] 
{\color{red}The series is said to converge if and only if the corresponding sequence is convergent, and if this is the case the limit of the sequence is the sum of the series.}


\end{enumerate}

\end{frame}

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\begin{frame}{2.2. Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}%\itemsep0.5em 

\item[8.] 
Applied to a series Cauchy's convergence test yields the following condition: 

\vspace{0.2cm}

%\item[9.] 
\begin{center}
\fbox{\parbox{10cm}
{\color{red}The series %(\ref{eq-15}) 
\begin{equation*}
a_1+a_2+\cdots+a_n+\cdots
%\label{eq-15}
\end{equation*}
converges if and only if to every $\varepsilon > 0$ there
exists an $n_0$ such that $$|a_n + a_{n+1} + \cdots + a_{n+p}| < \varepsilon $$ for all $n \ge n_0$ and $p \ge 0$. 
}}
\end{center}


\end{enumerate}

\end{frame}

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\begin{frame}{2.2. Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[9.] 
For $p = 0$ we find in particular that $|a_n| < \varepsilon$. 

\item[10.] 
Hence the general term of a convergent series tends to zero. 

\item[11.] 
This condition is necessary, but of course not sufficient.

\item[12.] 
If a finite number of the terms of the series (\ref{eq-15}) are omitted, the new series converges or diverges together with (\ref{eq-15}). 

\item[13.] 
In the case of convergence, let $R_n$ be the sum of the series which begins with the term $a_{n+1}$. 

\item[14.] 
Then the sum of the whole series is $S = s_n + R_n$. 


\end{enumerate}

\end{frame}

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\begin{frame}{2.2. Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[15.] 
The series (\ref{eq-15}) can be compared with the series 
\begin{equation}
|a_1| + |a_2| + \cdots + |a_n| + \cdots
\label{eq-16}
\end{equation}
formed by the absolute values of the terms. 

\item[16.] 
The sequence of partial sums of (\ref{eq-15}) is a contraction of the sequence corresponding to (\ref{eq-16}), for
\begin{equation*}
a_1 + a_2 + \cdots + a_n \le |a_1| + |a_2| + \cdots + |a_n| 
%\label{eq-16}
\end{equation*}

\item[17.] 
Therefore, convergence of (\ref{eq-16}) implies that the original series (\ref{eq-15}) is convergent. 

\item[18.] 
A series with the property that the series formed by the absolute values of the terms converges is said to be {\color{blue}absolutely convergent}.


\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. \hfill 作业2B-2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
Consider a sequence of functions $f_n(x)$, all defined on the same set $E$. 

\item[2.] 
If the sequence of values $\{f_n(x)\}$ converges for every $x$ that belongs to $E$, then the limit $f(x)$ is again a function on $E$. 

\item[3.] 
By definition, if $\varepsilon > 0$ and $x$ belongs to $E$ there exists an $n_0$ such that $|f_n(x) - f(x)| < \varepsilon$ for $n \ge n_0$, {\color{blue}but $n_0$ is allowed to depend on $x$}. 

\item[4.] 
{\color{red}
For instance, it is true that
$$
\lim\limits_{n\to\infty} \left(1 + \frac{1}{n} \right) x = x
$$
for all $x$, but in order to have $|(1+1/n)x-x| = |x|/n < \varepsilon$ for $n \ge n_0$ it is necessary that $n_0 > |x|/\varepsilon$. 
}


\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence.}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
Such an $n_0$ exists for every fixed $x$, {\color{blue}but the requirement cannot be met simultaneously for all $x$}.

\item[6.] 
We say in this situation that the sequence {\color{blue}converges pointwise}, but not {\color{blue}uniformly}. 

\item[7.] 
In positive formulation: 

\vspace{0.2cm}

\begin{center}
\fbox{\parbox{10cm}{
The sequence $\{f_n(x)\}$ converges uniformly to $f(x)$ on the set $E$ if to every $\varepsilon > 0$ there exists an $n_0$ such that $|f_n(x)- f(x)| <\varepsilon$ for all $n\ge n_0$ and all $x$ in $E$.
}}
\end{center}

\vspace{0.2cm}

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. \hfill 作业2B-3 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[8.] 
{\color{red}The most important consequence of uniform convergence is the following:

\begin{center}
\fbox{\parbox{10cm}{
The limit function of a uniformly convergent sequence of continuous functions is itself continuous. 
}}
\end{center}
}

\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[9.] 
Suppose that the functions $f_n(x)$ are continuous and tend uniformly to $f(x)$ on the set $E$. 

\item[10.] 
For any $\varepsilon > 0$ we are able to find an $n$ such that 
$|f_n(x) - f(x)| < \varepsilon/3$ for all $x$ in $E$. 

\item[11.] 
Let $x_0$ be a point in $E$. 

\item[12.] 
Because $f_n(x)$ is continuous at $x_0$ we can find $\delta > 0$ such that $|f_n(x) - f_n(x_0)| < \varepsilon/3$ for all $x$ in $E$ with $|x-x_0| < \delta$. 


\item[13.] 
Under the same condition on $x$ it follows that
\begin{equation*}
\begin{aligned}
|f(x) - f(x_0)| &\le |f(x) - f_n(x)| + |f_n(x)-f_n(x_0)| + |f_n(x_0)-f(x_0)| \\ 
& < \varepsilon,
\end{aligned}
%\label{eq-}
\end{equation*}
and we have proved that $f(x)$ is continuous at $x_0$.


\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[14.] 
{\color{blue}In the theory of analytic functions we shall find uniform convergence much more important than pointwise convergence.}

\item[15.] 
However, in most cases it will be found that the convergence is uniform only on a part of the set on which the functions are originally defined. 


\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[16.] 
Cauchy's necessary and sufficient condition has a counterpart for uniform convergence. We assert:

%\item[17.] 
\begin{center}
\fbox{\parbox{10cm}{\color{red}
The sequence $\{f_n(x)\}$ converges uniformly on $E$ if and only if to every $\varepsilon > 0$ there exists an $n_0$ such that $|f_m(x) - f_n(x)| < \varepsilon$ for all $m,n \ge n_0$ and all $x$ in $E$. 
}}
\end{center}


\item[17.] 
The necessity is again trivial. 

\item[18.] 
For the sufficiency we remark that the limit function $f(x)$ exists by the ordinary form of Cauchy's test. 

\item[19.] 
In the inequality $f_m(x) - f_n(x)| < \varepsilon$ we can keep $n$ fixed and let $m$ tend to $\infty$. 

\item[20.] 
It follows that $|f(x) - f_n(x)| \le \varepsilon$ for $n \ge n_0$ and all $x$ in $E$. 

\item[21.] 
Hence the convergence is uniform. 


\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[22.] 
For practical use the following test is the most applicable: 
%\item[24.] 

\begin{center}
\fbox{\parbox{10cm}{\color{red}
If a sequence of functions $\{f_n(x)\}$ is a contraction of a convergent sequence of
constants $\{a_n\}$, then the sequence $\{f_n(x)\}$ is uniformly convergent. 
}}
\end{center}

\item[23.] 
The hypothesis means that $|f_m(x) - f_n(x)| \le |a_m - a_n|$ on $E$, and the conlusion follows immediately by Cauchy's condition.

\item[24.] 
In the case of series this criterion, in a somewhat weaker form, becomes particularly simple. 


\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[25.] 
We say that a series with variable terms
\begin{equation*}
f_1(x) + f_2(x) + \cdots + f_n(x) + \cdots 
%\label{eq-}
\end{equation*}
has the series with positive terms
\begin{equation*}
a_1 + a_2 + \cdots + a_n + \cdots 
%\label{eq-}
\end{equation*}
for a {\color{blue}majorant} if it is true that $|f_n(x)|\le Ma_n$ for some constant $M$ and for all sufficiently large $n$; conversely, the first series is a {\color{blue}minorant} of the second. 

\item[26.] 
In these circumstances we have 
\begin{equation*}
|f_1(x) + f_2(x) + \cdots + f_n(x) + \cdots |
\le 
M(a_1 + a_2 + \cdots + a_n + \cdots )
%\label{eq-}
\end{equation*}

\item[27.] 
{\color{red}Therefore, if the majorant converges, the minorant converges uniformly. }

\item[28.] 
This condition is frequently referred to as the {\color{blue}Weierstrass M test}. 


\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[29.] 
It has the slight weakness that it applies only to series which are also absolutely convergent. 

\item[30.] 
The general principle of contraction is more complicated, but has a wider range of applicability.


\end{enumerate}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. Exercise - 1}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that a convergent sequence is bounded.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $\lim\limits_{n\to\infty} z_n = A$, prove that
$$
\lim\limits_{n\to\infty} \frac{1}{n} (z_1 + z_2 + \cdots + z_n)=A. 
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. Exercise - 3  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that the sum of an absolutely convergent series does not change if the terms are rearranged.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. Exercise - 4 \hfill 作业2B-4 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Discuss completely the convergence and uniform convergence of the sequence 
$\{nz^n\}_1^\infty$. 
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. Exercise - 5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Discuss the uniform convergence of the series
$$
\sum\limits_{n=1}^{\infty} \frac{x}{n(1+nx^2)}
$$
for real values of $x$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.3. Uniform Convergence. Exercise - 6}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $U = u_1 + u_2 +\cdots, V= v_1 + v_2 + \cdots$ are convergent series, prove that $UV = u_1v_1 + (u_1v_2 + u_2v_1) + (u_1v_3 + u_2v_2 + u_3v_1) + \cdots$ 
provided that at least one of the series is absolutely convergent. 
(It is easy if both series are absolutely convergent.
Try to arrange the proof so economically that the absolute convergence of the second series is not needed.)
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
A power series is of the form
\begin{equation}
a_0 + a_1z + a_2z^2 + \cdots + a_nz^n + \cdots 
\label{eq-17}
\end{equation}
where the coefficients $a_n$ and the variable $z$ are complex. 

\item[2.] 
A little more generally we may consider series
\begin{equation*}
\sum\limits_{n=0}^{\infty} a_n(z-z_0)^n
%\label{eq-}
\end{equation*}
which are power series with respect to the center $z_0$, but the difference is so
slight that we need not do so in a formal manner.

\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[3.] 
As an almost trivial example we consider the geometric series
$$
1 + z + z^2 + \cdots + z^n + \cdots
$$
whose partial sums can be written in the form
$$
1 + z + z^2 + \cdots + z^n = \frac{1-z^n}{1-z}.
$$

\item[4.] 
Since $z^n\to 0$ for $|z| < 1$ and $|z^n| \ge 1$ for $|z| \ge 1$ we conclude that the
geometric series converges to $1/(1-z)$ for $|z| < 1$, diverges for $|z|\ge l$.

\item[5.] 
{\color{blue}It turns out that the behavior of the geometric series is typical.} 

\item[6.] 
Indeed, we shall find that every power series converges inside a circle and diverges outside the same circle, except that it may happen that the series converges only for $z = 0$, or that it converges for all values of $z$.

\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. Theorem 2. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[7.] 
More precisely, we shall prove the following theorem due to Abel:

\item[8.] 
{\color{red} 
For every power series 
$$
a_0+a_1z+a_2z^2+\cdots+a_nz^n+\cdots
$$
there exists a number $R$, $0\le R\le\infty$, called the {\color{blue}radius of convergence}, with the following properties:
\begin{enumerate}
\item[(i)]  {\color{red} The series converges absolutely for every $z$ with $|z| < R$. 
If $0\le \rho < R$ the convergence is {\color{blue}uniform} for $|z|\le \rho$.}
\item[(ii)]  {\color{red} If $|z| > R$ the terms of the series are unbounded, and the series is consequently divergent.}
\item[(iii)]  {\color{red} In $|z| < R$ the sum of the series is an analytic function. The derivative can be obtained by termwise differentiation, and the derived series has the same radius of convergence.}
\end{enumerate}
}


\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. \hfill 作业2B-5 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[9.] 
The circle $|z| = R$ is called {\color{blue}the circle of convergence}; nothing is claimed about the convergence on the circle. 

\item[10.] 
We shall show that the assertions in the theorem are true if $R$ is chosen according to the formula
\begin{equation}
1/R = \underset{n\to\infty}{\mathrm{limsup}}\, \sqrt[n]{|a_n|}. 
\label{eq-18}
\end{equation}

\item[11.] 
This is known as {\color{blue}Hadamard's formula} for the radius of convergence.


\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[12.] 
If $|z| < R$ we can find $\rho$ so that $|z| < \rho < R$. 

\item[13.] 
Then $1/\rho > 1/R$, and by the definition of {\color{blue}limes superior} there exists an $n_0$ such that $|a_n|^{1/n} < 1/\rho$, $|a_n| < 1/\rho^n$ for $n \ge n_0$. 

\item[14.] 
It follows that $|a_nz^n| < (|z|/\rho)^n$ for large $n$, so that the power series (\ref{eq-17}) has a convergent geometric series as a majorant, and is consequently convergent. 


\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[15.] 
To prove the {\color{blue}uniform convergence} for $|z| \le \rho < R$ we choose a $\rho'$ with $\rho < \rho' < R$ and find $|a_nz^n| \le (\rho/\rho')^n$ for $n \ge n_0$. 

\item[16.] 
Since the {\color{blue}majorant} is convergent and has constant terms we conclude by {\color{blue}Weierstrass's M test} that the power series is uniformly convergent. 

\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[17.] 
If $|z| > R$ we choose $\rho$ so that $R < \rho < |z|$.

\item[18.] 
Since $1/\rho < 1/R$ there are arbitrarily large $n$ such that $|a_n|^{1/n} > 1/\rho$, $|a_n| > 1/\rho^n$. 

\item[19.] 
Thus $|a_nz^n| > (|z|/\rho)^n$ for infinitely many $n$, and the terms are unbounded.


\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[20.] 
The derived series $\sum\limits_1^\infty na_nz^{n-1}$ has the same radius of convergence, because $\sqrt[n]{n}\to 1$. 

\item[21.] 
Proof: Set $\sqrt[n]{n} = 1 + \delta_n$. Then $\delta_n > 0$, and by use of the binomial theorem $n = (1 + \delta_n)^n > 1 + \frac{1}{2}n(n-1)\delta_n^2$.

\item[22.] 
This gives $\delta_n^2 < 2/n$, and hence $\delta_n\to 0$.

\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[23.] 
For $|z| < R$ we shall write
\begin{equation*}
f(z) = \sum\limits_0^\infty a_nz^n = s_n(z) + R_n(z)
%\label{eq-}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
s_n(z) &= a_0 + a_1z + \cdots + a_{n-1}z^{n-1},\\ 
R_n(z) &= a_nz^n + a_{n+1}z^{n+1} + \cdots, 
\end{aligned}
%\label{eq-}
\end{equation*}
and also
\begin{equation*}
f_1(z) = \sum\limits_1^\infty na_nz^{n-1} = \lim\limits_{n\to\infty}s_n'(z). 
%\label{eq-}
\end{equation*}

\item[24.] 
We have to show that $f'(z) = f_1(z)$.

\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[25.] 
Consider the identity
\begin{equation}
\begin{aligned}
\frac{f(z)-f(z_0)}{z-z_0} - f_1(z_0) 
& = \left( \frac{s_n(z)-s_n(z_0)}{z-z_0} - s_n'(z_0) \right) \\
& +(s_n'(z_0) - f_1(z_0)) + \left( \frac{R_n(z)-R_n(z_0)}{z-z_0} \right) 
\end{aligned}
\label{eq-19}
\end{equation}
where we assume that $z \neq z_0$ and $|z|$, $|z_0| < \rho < R$.


\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[26.] 
The last term can be rewritten as
\begin{equation*}
\sum\limits_{k=n}^{\infty} a_k(z^{k-1} + z^{k-2}z_0 + \cdots + zz_0^{k-2} + z_0^{k-1}),
\end{equation*}
and we conclude that
\begin{equation*}
\left\vert \frac{R_n(z)-R_n(z_0)}{z-z_0} \right\vert 
\le \sum\limits_{k=n}^{\infty} k|a_k|\rho^{k-1}. 
\end{equation*}


\item[27.] 
The expression on the right is the remainder term in a convergent series. 


\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[28.] 
Hence we can find $n_0$ such that 
\begin{equation*}
\left\vert \frac{R_n(z)-R_n(z_0)}{z-z_0} \right\vert 
< \frac{\varepsilon}{3} 
\end{equation*}
for $n \ge n_0$. 

\item[29.] 
There is also an $n_1$ such that $|s_n'(z_0) - f_1(z_0)| < \varepsilon/3$ for $n\ge n_1$. 

\item[30.] 
Choose a fixed $n\ge n_0, n_1$.

\item[31.] 
By the definition of derivative we can find $\delta > 0$ such that $0 < |z-z_0| < \delta$ implies 
\begin{equation*}
\left\vert \frac{s_n(z)-s_n(z_0)}{z-z_0} - s_n'(z_0) \right\vert 
< \frac{\varepsilon}{3} 
\end{equation*}


\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[32.] 
When all these inequalities are combined it follows by (\ref{eq-19}) that
\begin{equation*}
\left\vert \frac{f(z)-f(z_0)}{z-z_0} - f_1(z_0) \right\vert < \varepsilon
\end{equation*}
when $0<|z-z_0|<\delta$. 

\item[33.] 
We have proved that $f'(z_0)$ exists and equals $f_1(z_0)$.

\item[34.] 
Since the reasoning can be repeated we have in reality proved much more.  


\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}



\item[35.] 
{\color{red}
A power series with positive radius of convergence has derivatives of all orders,}  and they are given explicitly by 
\begin{equation*}
\begin{aligned}
f(z) &= a_0 + a_1z + a_2z^2 + \cdots \\ 
f'(z) &= a_1 + 2a_2z + 3a_3z^2 + \cdots \\ 
f''(z) &= 2a_2 + 6a_3z + 12a_4z^2 + \cdots \\ 
\cdots & \cdots \\
f^{(k)}(z) &= k!a_k + \frac{(k+1)!}{1!}a_{k+1}z + \frac{(k+2)!}{2!}a_{k+2}z^2 + \cdots 
\end{aligned}
\end{equation*}


\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[36.] 
In particular, if we look at the last line we see that $a_k = f^{(k)}(0)/k!$, and
the power series becomes
\begin{equation*}
f(z) = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \cdots + \frac{f^{(n)}(0)}{n!} z^n +\cdots .
\end{equation*}

\item[37.] 
This is the familiar {\color{blue}Taylor-Maclaurin development}, but we have proved it {\color{orange}only under the assumption that $f(z)$ has a power series development.} 

\item[38.] 
We do know that the development is uniquely determined, if it exists, {\color{orange}but the main part is still missing, namely that every analytic function has a Taylor development.} 

\end{enumerate}

\end{frame}

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\begin{frame}{2.4. Power Series. Exercise - 1 \hfill 作业2B-6 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Expand $(1-z)^{-m}$, $m$ a positive integer, in powers of $z$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. Power Series. Exercise - 2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Expand $\frac{2z+3}{z+1}$ in powers of $z-1$. What is the radius of convergence?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. Power Series. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the radius of convergence of the following power series:
$$
\sum n^pz^n, \hspace{0.3cm}
\sum \frac{z^n}{n!}, \hspace{0.3cm}
\sum n!z^n, \hspace{0.3cm}
\sum q^{n^2}z^n, \,\,(|q|<1)\hspace{0.3cm}
\sum z^{n!}.
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. Power Series. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $\sum a_nz^n$ has radius of convergence $R$, what is the radius of convergence of 
$\sum a_nz^{2n}$? of $\sum a_n^2z^n$?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. Power Series. Exercise - 5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $f(z) = \sum a_nz^n$, what is $\sum n^3a_nz^n$?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. Power Series. Exercise - 6}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $\sum a_nz^n$ and $\sum b_nz^n$ have radii of convergence $R_1$ and $R_2$, show that the radius of convergence of $\sum a_nb_nz^n$ is at least $R_1R_2$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. Power Series. Exercise - 7 \hfill 作业 2B-7 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $\lim\limits_{n\to\infty} \frac{|a_n|}{|a_{n+1}|} = R$, prove that $\sum a_nz^n$ has radius of convergence $R$.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. Power Series. Exercise - 8}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
For what values of $z$ is 
$$
\sum\limits_{0}^{\infty} \left( \frac{z}{1+z} \right)^n
$$
convergent?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.4. Power Series. Exercise - 9}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Same question for
$$
\sum\limits_{0}^{\infty} \frac{z^n}{1+z^{2n}}. 
$$

}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{2.5. Abel's Limit Theorem. Theorem 3. \hfill 作业 2B-8 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
There is a second theorem of Abel's which refers to the case where a power series converges at a point of the circle of convergence. 

\item[2.] 
We lose no generality by assuming that $R = 1$ and that the convergence takes place at $z = 1$.

\item[3.] 
{\color{red}Theorem 3. If $\sum\limits_{0}^{\infty}a_n$ converges, then $f(z) = \sum\limits_{0}^{\infty}a_nz^n$ tends to $f(1)$ as $z$ approaches 1 in such a way that $\frac{|1-z|}{1-|z|}$ remains bounded. }

\item[4.] 
Geometrically, the condition means that $z$ stays in an angle $< 180°$ with vertex $1$, symmetrically to the part $(-\infty, 1)$ of the real axis. 

\item[5.] 
It is customary to say that the approach takes place in a {\color{blue}Stolz angle}.

\end{enumerate}

\end{frame}

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\begin{frame}{2.5. Abel's Limit Theorem. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[6.] 
Proof.
We may assume that $\sum\limits_{0}^{\infty}a_n = 0$, for this can be attained by adding a constant to $a_0$. 

\item[7.] 
We write $s_n = a_0 + a_1 + \cdots + a_n$ and make use of the identity ({\color{blue}summation by parts}) 
\begin{equation*}
\begin{aligned}
s_n(z) &= a_0 + a_1z + \cdots + a_nz^n \\ 
& = s_0 + (s_1-s_0)z + \cdots + (s_n - s_{n-1})z^n \\ 
& = s_0(1 - z) + s_1(z - z^2) + \cdots + s_{n-1}(z^{n-1} - z^n) + s_nz^n \\ 
& = (1 - z)(s_0 + s_1z + \cdots + s_{n-l}z^{n-1}) + s_nz^n.
\end{aligned}
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{frame}{2.5. Abel's Limit Theorem. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[8.] 
But $s_nz^n \to 0$, so we obtain the representation
\begin{equation*}
f(z) = (1-z)\sum\limits_{0}^{\infty} s_nz^n. 
\end{equation*}

\item[9.] 
{\color{blue}We are assuming that $|1 -z| \le K(1 -|z|)$, say, and that $s_n\to 0$. } 

\item[10.] 
Choose $m$ so large that $|s_n| < \varepsilon$ for $n \ge m$. 

\item[11.] 
The remainder of the series $\sum s_nz^n$, from $n = m$ on, is then dominated by the geometric series
\begin{equation*}
\varepsilon \sum\limits_{m}^{\infty}|z|^n = 
\varepsilon |z|^m/(1-|z|) < \varepsilon /(1-|z|).
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{frame}{2.5. Abel's Limit Theorem. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[12.] 
It follows that
\begin{equation*}
|f(z)| \le |1 - z| \left\vert \sum\limits_{0}^{m-1} s_kz^k \right\vert + K\varepsilon.
\end{equation*}

\item[13.] 
The first term on the right can be made arbitrarily small by choosing $z$ sufficiently close to 1, and we conclude that $f(z)\to 0$ when $z \to 1$ subject to
the stated restriction.

\end{enumerate}

\end{frame}


\end{document}

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